Problems

An equilateral triangle \(ABC\) has sides of length \(a\). The points \(P\), \(Q\) and \(R\) lie on the sides \(BC\), \(CA\) and \(AB\), respectively, such that the length \(BP\) is \(x\) and \(QR\) is parallel to \(CB\). Show that \[ (\sqrt{3}\cot\phi + 1)(\sqrt{3}\cot\theta + 1)x = 4(a - x), \] where \(\theta = \angle CPQ\) and \(\phi = \angle BRP\). A horizontal triangular frame with sides of length \(a\) and vertices \(A\), \(B\) and \(C\) is fixed on a smooth horizontal table. A small ball is placed at a point \(P\) inside the frame, in contact with side \(BC\) at a distance \(x\) from \(B\). It is struck so that it moves round the triangle \(PQR\) described above, bouncing off the frame at \(Q\) and then \(R\) before returning to point \(P\). The frame is smooth and the coefficient of restitution between the ball and the frame is \(e\). Show that \[ x = \frac{ae}{1 + e}. \] Show further that if the ball continues to move round \(PQR\) after returning to \(P\), then \(e = 1\).

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Solution:

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\begin{align*} && \frac{x}{\sin \phi} &= \frac{PR}{\sin 60^{\circ}} \\ && \frac{a-x}{\sin (120^{\circ}-\theta)} &= \frac{QP}{\sin 60^{\circ}} \\ && \frac{PR}{\sin \theta} &= \frac{QP}{\sin(120^{\circ}-\phi)} \\ \\ \Rightarrow && PR &= \frac{\sqrt3}2 \frac{x}{\sin \phi} \\ && QP &= \frac{\sqrt3}2 \frac{a-x}{\sin(120^{\circ}-\theta)} \\ \Rightarrow && \frac{\sqrt3}2 \frac{x}{\sin \phi} \frac{1}{\sin \theta} &= \frac{\sqrt3}2 \frac{a-x}{\sin(120^{\circ}-\theta)} \frac{1}{\sin(120^{\circ}-\phi)} \\ \Rightarrow && a-x &= \frac{(\frac{\sqrt3}2 \cos \phi + \frac12 \sin \phi)(\frac{\sqrt3}2 \cos \theta + \frac12 \sin \theta)}{\sin \phi \sin \theta} x \\ \Rightarrow && 4(a-x) &= (\sqrt3 \cot \phi + 1)(\sqrt 3 \cot \theta + 1) x \end{align*}

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Notice that \(e = \frac{\tan 60^{\circ}}{\tan (120^{\circ}-\theta)} = \frac{\tan \phi}{\tan 60^{\circ}}\) or \begin{align*} && \tan \phi &= \sqrt 3 e \\ && \frac{\sqrt3}{e} &= \tan(120^{\circ} - \theta) \\ &&&= \frac{-\sqrt3 - \tan \theta}{1 - \sqrt3 \tan \theta} \\ &&&= \frac{\sqrt3 \cot \theta +1}{\sqrt3-\cot \theta} \\ \Rightarrow && \sqrt3(-e-1)\cot \theta &= e-3 \\ \Rightarrow && \cot\theta &= \frac{3-e}{\sqrt{3}(1+e)} \\ \\ \Rightarrow && 4(a-x) &= (\sqrt3 \cot \phi + 1)(\sqrt 3 \cot \theta + 1) x \\ &&&= \left (\sqrt3 \frac{1}{\sqrt3 e} + 1 \right) \left (\sqrt 3\frac{3-e}{\sqrt{3}(1+e)}+1 \right) x \\ &&&= \frac{1+e}{e}\frac{3-e+1+e}{1+e} x \\ \Rightarrow && (a-x) &= \frac{1}{e}x \\ \Rightarrow && a &= \frac{1+e}{e}x \\ \Rightarrow && x &= \frac{ae}{1+e} \end{align*} The ball will continue to move around \(PQR\) if \(e \tan(120^{\circ} - \phi) = \tan \theta\) ie \begin{align*} && e \frac{-\sqrt3-\tan \phi}{1-\sqrt3 \tan \phi} &= \tan \theta \\ \Rightarrow && e \frac{\sqrt3 + \sqrt3 e}{3e-1} &= \frac{\sqrt3(1+e)}{3-e} \\ \Rightarrow && \frac{e}{3e-1} &= \frac{1}{3-e} \tag{\(e \neq -1\)} \\ \Rightarrow && 3e-e^2 &= 3e-1 \\ \Rightarrow && e^2 &= 1 \\ \Rightarrow && e &= 1 \end{align*}

A prison consists of a square courtyard of side \(b\) bounded by a perimeter wall and a square building of side \(a\) placed centrally within the courtyard. The sides of the building are parallel to the perimeter walls. Guards can stand either at the middle of a perimeter wall or in a corner of the courtyard. If the guards wish to see as great a length of the perimeter wall as possible, determine which of these positions is preferable. You should consider separately the cases \(b<3a\) and \(b>3a\,\).

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The orange guard will always see \(2b+b-a = 3b-a\) The blue guard will see \(b + \frac{b(b-a)}{a} = \frac{b^2}{a}\) if \(b < 3a\) and \(3b + \frac{b(b-3a)}{(b-a)} = \frac{2b(2b-3a)}{b-a}\). Therefore the blue guard always sees more if \(b > 3a\). He sees more in the other case if \begin{align*} && \frac{b^2}{a} &> 3b - a \\ \Leftrightarrow && \frac{b^2}{a^2} &> 3\frac{b}{a} - 1 \\ \Leftrightarrow && x^2 - 3x + 1 &> 0\\ \Leftrightarrow && x > \frac{3 + \sqrt{5}}{2} \text{ or } x < \frac{3-\sqrt{5}}{2} \end{align*} Since \(b > a\) we must have \(b > \frac{3+\sqrt{5}}2 a\) There is an alternative interpretation which is that the orange guard is in the top left corner, ie

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In this case the green guard will always see \(2b + \frac{2b(b-a)}{b+a} = \frac{4b^2}{b+a}\) Comparing \(\frac{4b^2}{b+a}\) with \(\frac{b^2}{a}\) we can see the former is larger if \(3a > b\). Comparing \(\frac{4b^2}{b+a}\) with $$

A hollow circular cylinder of internal radius \(r\) is held fixed with its axis horizontal. A uniform rod of length \(2a\) (where \(a < r\)) rests in equilibrium inside the cylinder inclined at an angle of \(\theta\) to the horizontal, where \(\theta\ne0\). The vertical plane containing the rod is perpendicular to the axis of the cylinder. The coefficient of friction between the cylinder and each end of the rod is \(\mu\), where \(\mu > 0\). Show that, if the rod is on the point of slipping, then the normal reactions \(R_1\) and \(R_2\) of the lower and higher ends of the rod, respectively, on the cylinder are related by \[ \mu(R_1+R_2) = (R_1-R_2)\tan\phi \] where \(\phi\) is the angle between the rod and the radius to an end of the rod. Show further that \[ \tan\theta = \frac {\mu r^2}{r^2 -a^2(1+\mu^2)}\,. \] Deduce that \(\lambda <\phi \), where \(\tan\lambda =\mu\).

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Solution:

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Let \(M\) be the midpoint of \(AB\), then \begin{align*} \overset{\curvearrowright}{M}: && R_1 \sin \phi-\mu R_1 \cos \phi &= R_2 \sin \phi+\mu R_2 \cos \phi \\ \Rightarrow && (R_1-R_2) \tan \phi &= \mu(R_1+R_2) \end{align*} As required. \begin{align*} && \cos \phi = \frac{a}{r} &,\,\, \sin \phi = \frac{\sqrt{r^2-a^2}}{r} \\ \text{N2}(\rightarrow): && R_1\cos(\phi + \theta)+\mu R_1 \sin(\phi + \theta) &= R_2 \cos(\theta - \phi) + \mu R_2 \sin(\theta - \phi) \\ \Rightarrow && R_1(\cos \theta \cos \phi - \sin \theta \sin \phi)+ \mu R_1 (\sin \theta \cos \phi + \cos \theta \sin \phi) &= R_2 (\cos\theta \cos \phi + \sin \theta \sin \phi)+ \mu R_2 (\sin \theta \cos \phi - \cos \theta \sin \phi) \\ && R_1 (1 - \tan \theta \tan \phi)+\mu R_1 (\tan \theta + \tan \phi) &= R_2(1 + \tan \theta \tan \phi) +\mu R_2 (\tan \theta - \tan \phi) \\ && 0 &= (R_1-R_2)(1+\mu \tan \theta)+(R_1+R_2)(-\tan \theta \tan\phi+\mu \tan \phi) \\ \Rightarrow && \frac{R_1+R_2}{R_1-R_2} &= \frac{1+\mu \tan \theta}{\tan \phi (\tan \theta - \mu))} \\ \Rightarrow && \frac{\tan \phi}{\mu} &= \frac{1+\mu \tan \theta}{\tan \phi (\tan \theta - \mu))} \\ \Rightarrow && \tan^2 \phi &= \frac{\mu(1+\mu \tan \theta)}{\tan \theta - \mu} \\ \Rightarrow && \frac{r^2-a^2}{a^2} &= \frac{\mu(1+\mu \tan \theta)}{\tan \theta - \mu} \\ \Rightarrow && \tan \theta (r^2-a^2-a^2\mu^2) &= \mu a^2+\mu(r^2-a^2) \\ \Rightarrow && \tan \theta &= \frac{\mu r^2}{r^2-(1+\mu^2)a^2} \end{align*} Since \(\mu r^2 > 0\) we must also have \(r^2 > a^2(1+\mu^2)\) ie \(\\sec^2 \phi > 1 + \mu^2 = \sec^2 \lambda\) and the result follows.

A thin uniform circular disc of radius \(a\) and mass \(m\) is held in equilibrium in a horizontal plane a distance \(b\) below a horizontal ceiling, where \(b>2a\). It is held in this way by \(n\) light inextensible vertical strings, each of length \(b\); one end of each string is attached to the edge of the disc and the other end is attached to a point on the ceiling. The strings are equally spaced around the edge of the disc. One of the strings is attached to the point \(P\) on the disc which has coordinates \((a,0,-b)\) with respect to cartesian axes with origin on the ceiling directly above the centre of the disc. The disc is then rotated through an angle \(\theta\) (where \(\theta<\pi\)) about its vertical axis of symmetry and held at rest by a couple acting in the plane of the disc. Show that the string attached to~\(P\) now makes an angle \(\phi\) with the vertical, where \[ b\sin\phi = 2a \sin\tfrac12 \theta\,. \] Show further that the magnitude of the couple is \[ \frac {mga^2\sin\theta}{\sqrt{b^2-4a^2\sin^2 \frac12\theta \ } \ }\,. \] The disc is now released from rest. Show that its angular speed, \(\omega\), when the strings are vertical is given by \[ \frac{a^2\omega^2}{4g} = b-\sqrt{b^2 - 4a^2\sin^2 \tfrac12\theta \;}\,. \]

The lengths of the sides of a rectangular billiards table \(ABCD\) are given by \(AB = DC = a\) and \(AD=BC = 2b\). There are small pockets at the midpoints \(M\) and \(N\) of the sides \(AD\) and \(BC\), respectively. The sides of the table may be taken as smooth vertical walls. A small ball is projected along the table from the corner \(A\). It strikes the side \(BC\) at \(X\), then the side \(DC\) at \(Y\) and then goes directly into the pocket at \(M\). The angles \(BAX\), \(CXY\) and \(DY\!M\) are \(\alpha\), \(\beta\) and \(\gamma\) respectively. On each stage of its path, the ball moves with constant speed in a straight line, the speeds being \(u\), \(v\) and \(w\) respectively. The coefficient of restitution between the ball and the sides is \(e\), where \(e>0\).

1. Show that \(\tan\alpha \tan \beta = e\) and find \(\gamma\) in terms of \(\alpha\).
2. Show that \(\displaystyle \tan\alpha = \frac {(1+2e)b} {(1+e)a}\) and deduce that the shot is possible whatever the value of \(e\).
3. Find an expression in terms of \(e\) for the fraction of the kinetic energy of the ball that is lost during the motion.

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Solution:

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1. The initial velocity is \(u = \binom{u \cos\alpha}{u \sin \alpha}\), therefore \(v = \binom{v_x}{u \sin \alpha}\). Newton's experimental law tells us \(v_x = -e u_x = -eu \cos\alpha\), therefore \(v = \binom{-eu \cos \alpha}{u \sin \alpha} = \binom{-v \sin \beta}{v\cos \beta} \Rightarrow -\tan \beta = -e \cot \alpha \Rightarrow \tan \alpha \tan \beta = e\). There is nothing special about the result here, and so it must also be the case that \(\tan \beta \tan \gamma = e \Rightarrow \tan \gamma = \tan \alpha\)
2. \(\tan \alpha = \frac{XB}{BA}\) so the point \(X\) is at \((a, \tan \alpha a)\). The point \(Y\) satisfies \(\tan \beta = \frac{CY}{CX} = \frac{CY}{2b - \tan \alpha a}\) so the point \(Y\) is \((a-(2b - a \tan \alpha)\tan \beta,2b) = (a - 2b\tan \beta + ea, 2b) = ((1+e)a-2b\tan \beta, 2b)\). Finally, the point \(M\) is the midpoint, so \(\tan \gamma = \frac{DM}{DY}\) so \(M\) is the point \((0, 2b - ((1+e)a-2b\tan \beta)\tan \gamma) = (0, 2b - (1+e)a \tan \gamma - 2b e) = (0, (2b(1-e) - (1+e)a \tan \gamma)\), but \(M\) is the point \((0, b)\), ie \begin{align*} && b &= 2b(1-e) - (1+e)a \tan \gamma \\ \Rightarrow && b+2eb &= (1+e)a \tan \gamma \\ \Rightarrow && \tan \gamma &= \frac{(1+2e)b}{(1+e)a} \\ \Rightarrow && \tan \alpha &= \frac{(1+2e)b}{(1+e)a} \end{align*} Since \( \frac{(1+2e)b}{(1+e)a} = \frac{b}{a} + \frac{e}{1+e}b\) which is clearly an increasing function of \(e\) on \([0,1]\), so \(\tan \alpha \in \left [\frac{b}{a}, \frac{3b}{2a} \right]\) which are all all angles which place \(X\) in sensible places, therefore we can always hit the middle pocket. (Except \(e = 0\), where we would put the ball in \(N\), but we are given \(e > 0\)).
3. After the first collision the velocity is \(\binom{-eu \cos \alpha}{u \sin \alpha}\) after the second collision the velocity is \(\binom{-eu \cos \alpha}{-eu \sin \alpha}\). Initial kinetic energy is therefore \(\frac12 m u^2\) and final kinetic energy is \(\frac12 m e^2u^2\) therefore the fraction lost is \(\frac{\frac12 m u^2 - \frac12 m e^2u^2}{\frac12 m u^2} = 1-e^2\)

A smooth cylinder with circular cross-section of radius \(a\) is held with its axis horizontal. A~light elastic band of unstretched length \(2\pi a\) and modulus of elasticity \(\lambda\) is wrapped round the circumference of the cylinder, so that it forms a circle in a plane perpendicular to the axis of the cylinder. A particle of mass \(m\) is then attached to the rubber band at its lowest point and released from rest.

1. Given that the particle falls to a distance \(2a\) below the below the axis of the cylinder, but no further, show that \[ \lambda = \frac{9\pi m g}{(3\sqrt3-\pi)^2} \;. \]
2. Given instead that the particle reaches its maximum speed at a distance \(2a\) below the axis of the cylinder, find a similar expression for \(\lambda\)\,.

A light rod of length \(2a\) is hung from a point \(O\) by two light inextensible strings \(OA\) and \(OB\) each of length \(b\) and each fixed at \(O\). A particle of mass \(m\) is attached to the end \(A\) and a particle of mass \(2m\) is attached to the end \(B.\) Show that, in equilibrium, the angle \(\theta\) that the rod makes the horizontal satisfies the equation \[ \tan\theta=\frac{a}{3\sqrt{b^{2}-a^{2}}}. \] Express the tension in the string \(AO\) in terms of \(m,g,a\) and \(b\).

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Solution:

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The centre of mass of the rod will be at a point \(G\) which divides the rod in a ratio \(1:2\). Let \(M\) be the midpoint of \(AB\), so \(|AM| = a\) To be in equilibrium \(G\) must lie directly below \(O\). Note that \(OM^2 +a^2 = b^2 \Rightarrow OM = \sqrt{b^2-a^2}\) Notice that \(AG = \frac{4}{3}a\) and \(AM = a\), so \(|MG| = \frac13 a\). Therefore \(\displaystyle \tan \theta = \frac{\tfrac{a}{3}}{\sqrt{b^2-a^2}} \Rightarrow \tan \theta = \frac{a}{3\sqrt{b^2-a^2}}\).

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Notice that \(\frac{\sin \beta}{\frac43a} = \frac{\sin \angle OGA}{b} = \frac{\sin \alpha}{\frac23 a} \Rightarrow \sin \beta = 2 \sin \alpha\) \begin{align*} \text{N2}(\rightarrow, A): && C\cos \theta - T_A \sin \beta&= 0 \\ \text{N2}(\rightarrow, B): && T_B \sin \alpha - C\cos \theta &= 0 \\ \Rightarrow && T_B \sin \alpha &= T_A\sin \beta \\ \Rightarrow && T_B &= 2T_A \\ \text{N2}(\uparrow, A): && T_A \cos \beta+C\sin \theta-mg &= 0 \\ \text{N2}(\uparrow, B): && T_B \cos \alpha- C\sin \theta-2mg &= 0 \\ \Rightarrow && T_A \cos \beta+2T_A \cos \alpha&=3mg \\ \end{align*} Using the cosine rule: \((\frac23a)^2 = b^2 + OG^2 - 2b|OG|\cos \alpha\) and \((\frac43a)^2 = b^2 + OG^2-2b|OG|\cos \beta\). \(|OG|^2 = b^2 + (\frac43a)^2-\frac83ab \cos \angle A = b^2 +\frac{16}{9}a^2-\frac83a^2 = b^2-\frac89a^2\). Therefore \(\cos \alpha = \frac{2b^2-\frac43a^2}{2b |OG|}\), \(\cos \beta = \frac{2b^2-\frac83a^2}{2b|OG|}\) Therefore \(\cos \beta + 2 \cos \alpha = \frac{18b^2-16a^2}{6b|OG|} = \frac{9b^2-8a^2}{b\sqrt{9b^2-a^2}} = \frac{\sqrt{9b^2-a^2}}b\) Therefore \(\displaystyle T_A = \frac{3bmg}{\sqrt{9b^2-8a^2}}\)

\(\ \)

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1. as small as possible,
2. as large as possible.

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Solution:

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The area he cannot see is \begin{align*} &&A &= \underbrace{8^2}_{\text{everywhere above}(4,4)} - \underbrace{4^2}_{\text{inner square}} - \underbrace{\frac12 \cdot 4 \cdot (\frac32(4-p)+p - 4)}_{\text{blue triangle}} - \underbrace{\frac12 \cdot 4 \cdot \frac{4(12-p)}{8-p}}_{\text{green triangle}} \\ &&&= 48 - 3(4-p)-2(p-4) - \frac{8(12-p)}{8-p} \\ &&&= 36-5p-\frac{32}{8-p} \\ \\ \Rightarrow && \frac{\d A}{\d p} &= -5 + \frac{32}{(8-p)^2} \\ &&&> 0 \text{ if } 0 \leq p \leq 4 \end{align*}

1. Since the area not visible is increasing as \(p\) increases, we would like \(p\) to be as large as possible, ie \(p = 4\).
2. Similarly, he can see the most when \(p =0\)

A smooth uniform sphere, with centre \(A\), radius \(2a\) and mass \(3m,\) is suspended from a fixed point \(O\) by means of a light inextensible string, of length \(3a,\) attached to its surface at \(C\). A second smooth unifom sphere, with centre \(B,\) radius \(3a\) and mass \(25m,\) is held with its surface touching \(O\) and with \(OB\) horizontal. The second sphere is released from rest, falls and strikes the first sphere. The coefficient of restitution between the spheres is \(3/4.\) Find the speed \(U\) of \(A\) immediately after the impact in terms of the speed \(V\) of \(B\) immediately before impact. The same system is now set up with a light rigid rod replacing the string and rigidly attached to the sphere so that \(OCA\) is a straight line. The rod can turn freely about \(O\). The sphere with centre \(B\) is dropped as before. Show that the speed of \(A\) immediately after impact is \(125U/127.\)

A rubber band band of length \(2\pi\) and modulus of elasticity \(\lambda\) encircles a smooth cylinder of unit radius, whose axis is horizontal. A particle of mass \(m\) is attached to the lowest point of the band, and hangs in equilibrium at a distance \(x\) below the axis of the cylinder. Obtain an expression in terms of \(x\) for the stretched length of the band in equilibrium. What is the value of \(\lambda\) if \(x=2\)?

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If \(\alpha\) is as labelled then \(\cos \alpha = \frac{1}{x}, \sin \alpha = \frac{\sqrt{x^2-1}}{x}, \tan \alpha = \sqrt{x^2-1}\). We also have the full length of the rubber band is \(2\pi - 2\alpha +2\tan \alpha\) so the extension is \(2 \l \sqrt{x^2-1} - \cos^{-1} \l \frac{1}{x}\r \r\) Therefore \(T = \frac{\l \sqrt{x^2-1} - \cos^{-1} \l \frac{1}{x}\r \r\lambda}{\pi}\). If \(x = 2\), \(T = \frac{\sqrt{3} - \frac{\pi}{3}}{\pi} \lambda, \sin \alpha = \frac{\sqrt{3}}{2}\) \begin{align*} \text{N2}(\uparrow): && 2T\sin \alpha - mg &= 0 \\ \Rightarrow && \frac{\sqrt{3} - \frac{\pi}{3}}{\pi} \lambda \sqrt{3} &= mg \\ \Rightarrow && \lambda &= \frac{\sqrt{3}\pi}{(3\sqrt{3}-\pi)}mg \end{align*}

Ice snooker is played on a rectangular horizontal table, of length \(L\) and width \(B\), on which a small disc (the puck) slides without friction. The table is bounded by smooth vertical walls (the cushions) and the coefficient of restitution between the puck and any cushion is \(e\). If the puck is hit so that it bounces off two adjacent cushions, show that its final path (after two bounces) is parallel to its original path. The puck rests against the cushion at a point which divides the side of length \(L\) in the ratio \(z:1\). Show that it is possible, whatever \(z\), to hit the puck so that it bounces off the three other cushions in succession clockwise and returns to the spot at which it started. By considering these paths as \(z\) varies, explain briefly why there are two different ways in which, starting at any point away from the cushions, it is possible to perform a shot in which the puck bounces off all four cushions in succession clockwise and returns to its starting point.

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Solution:

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The puck sets off at some velocity \(\displaystyle \binom{u_x}{u_y}\), after the first bounce off the wall parallel to the \(y\)-axis, it has velocity \(\displaystyle \binom{-eu_x}{u_y}\). After it bounces off the wall parallel to the \(x\)-axis, it has velocity \(\displaystyle \binom{-eu_x}{-eu_y}\) which is clearly parallel to the original velocity.

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If the puck bounces off 3 walls and returns to the same point the shape formed must be a parallelogram. We need to hit the point on the opposite side which is in a ratio of \(1:z\), but this must be possible if we aim towards the side further away.

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For a fixed path, as \(z\) increases we generate more parallelograms which cross ours (on two of the legs) twice. As they move the full length it will cover the full leg of the parallogram. Similarly going the other way will cover the other leg of the parallelogram. Therefore from every point there are two circuits round the table